Talk:Nemytskii operator

Latest comment: 10 years ago by Jjh1993

In the section of the Boundless Theorem it may be added, that there exists a more general form where one concludes from , , that the Nemytskii operator is a bounded and continuous map from to .

This may belong to another article; but I have not found a better place yet.

Jjh1993 (talk) 11:16, 8 November 2014 (UTC)Reply